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When Quantum Mechanics is thrashed by non-physicists #1

  1. Jan 4, 2015 #1

    dextercioby

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    Historically, we know that above any other discipline of physics, quantum mechanics has attracted other types of scientists, such as mathematicians, chemists, specialists in information technology, but also, philosophers of science. While most of their historical contributions are really valuable (think von Neumann or Weyl), it is with considerable regret that I sense that nowadays, in the huge sea of 'which interpretation is better' viewpoints, their contributions are rather distructive, (perhaps I exaggerate here) or at least very challenging.
    I recently bumped into an article on the preprints server:

    http://arxiv.org/abs/1412.2701v1

    thrashing the idea that a (unit) vector of a Hilbert space can represent the state of a physical quantum system. One of the authors is a philosopher, the other a mathematician and. surprisingly, throughout the paper the mathematics of finite-dim. vector spaces (known to be improper for QM) is used. I don't think, however, that the lack of rigor in maths can be the point which turns their paper from a correct one into a wrong one.

    Have a read of it, please, and tell me where they go wrong, if anywhere. (As a joking sidepoint: Is there a catholic formalism of QM, too, because the literature is flooded with protestant ones challenging the orthodox one??:D)

    Thanks,
    :)
     
  2. jcsd
  3. Jan 4, 2015 #2

    ShayanJ

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    I don't think I know enough physics to comment on the technical parts. But I should say I don't accept that such papers are "destructive" or anything that bad! But "challenging", I accept. But I'm surprised that you're using this word as it is a weaker form of destructive! Of course if physicists overcome such challenges, it doesn't mean a waste of time or effort because they surely gained a deeper understanding. On the other hand, if physicists can't come up with an answer, then it means the those challenges should themselves point to a deeper thing.
    But I suspect you mean the authors have a misunderstanding about QM. Which I don't think is probable!

    Here I like to quote some part of Feynman's Nobel lecture:
     
  4. Jan 4, 2015 #3

    bhobba

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    Had a quick squiz.

    Am I imagining something or is the opening line totally 'whaco':
    'In this paper we derive a theorem which proves that the physical interpretation implied by the first postulate of quantum mechanics (QM) is inconsistent with the orthodox formalism.'

    How can a postulate be inconsistent with the formalism it is part of?

    I am not too worried about the finite dimensional thing because I view QM in the Rigged Hilbert Space formalism and think of finite dimensional states as the physical ones while the usual ones we work with are from the dual of all finite dimensional vectors for mathematical convenience. Of course you cant express the laws of QM without it so my view is a bit contrived.

    Thanks
    Bill
     
    Last edited: Jan 4, 2015
  5. Jan 4, 2015 #4

    Matterwave

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    I believe they are saying the "physical interpretation" of the first postulate is what contradicts the orthodox formalism, and not the first postulate itself. I don't know if they mean the first postulate is inconsistent with any physical interpretation or simply the one we ascribe to it via the orthodox formalism (which, iirc, is not very much at all, unless they mean the Copenhagen interpretation as the "orthodox formalism"?).

    I have only read the abstract by the way, as I am currently jet-lagged and in no mood to read a philosophy paper on the interpretations of the quantum state.
     
  6. Jan 4, 2015 #5

    atyy

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    The claim of the paper seems rather strong to me, and I don't know whether it is right. But in the context of relativity, I think I can supply an example in the same spirit. The quantum state is defined using a plane of simultaneity. If we only accept invariant quantities as real in relativity, then the quantum state is not real, since simultaneity is not absolute (without specification of a family of observers) in relativity.

    Also, when there are sequential measurements, the quantum dynamics includes wave function collapse. The state is Lorentz covariant under changes of inertial frame if time evolution is unitary, but not if collapse is also included. Fortunately, localized events like measurement outcomes and their probabilities are invariant, so those can be considered real, and there is no problem with quantum mechanics and relativity.

    Of course, if one removed invariance as a requirement for the physical state, one could have an aether frame, and the quantum state could be FAPP real in that frame.
     
  7. Jan 4, 2015 #6

    Matterwave

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    If the Authors' gripe with QM is that it is not Lorentz covariant, then certainly that is not a real gripe? QM is obviously not Lorentz covariant, it was never formulated to be. That's where QFT comes in...
     
  8. Jan 4, 2015 #7

    atyy

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    I'm including QFT when I say QM. I think they require real things to be invariant, but the state is defined using simultaneity, and usually we don't consider simultaneity to be invariant.

    My remark about collapse is an additional point. In QFT (let's say the Schroedinger functional picture) the state evolution is covariant only for unitary evolution, but not if collapse is included. So the state dynamics is not even covariant (let alone invariant).
     
  9. Jan 5, 2015 #8

    vanhees71

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    First of all: Weyl and von Neumann were mathematicians, not philosophers. Both were partially wrong with the physics part but have helped a lot to understand the mathematical foundations of quantum theory (and in the case of Weyl to General Relativity).

    I don't know about the present paper, but when I read the abstract, I'm inclined not to read further, because nobody claims that the quantum state is represented by a Hilbert-space vector. It's very easy to disprove this idea. The most simple argument against this is simply the existence of half-integer spins of particles. It's not Hilbert-space vectors that represent a state but rays in Hilbert space or, equivalently for pure states and including the more general concept of mixed states, statistical operators, i.e., positive semidefinite self-adjoint operators with trace 1.
     
  10. Jan 5, 2015 #9
    Maybe you should read section 4 of the paper, Ithink their argument is independent of the formal distinction between rays or statistical operators and vectors in complex Hilbert space.
     
  11. Jan 5, 2015 #10

    Demystifier

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    I have read the paper and I think I understood it. In my opinion, they are not wrong. This is because their claims are not really so radical as you might think they are. To understand that, one needs to understand carefully what they really mean by verbal expressions such as "physical state".

    First, even though they prove their theorems in a finite dimensional space, that's not a problem. Their theorems are akin to the celebrated Kochen-Specker theorem, which is also proved in a finite dimensional space, while nobody considers this to be a problem for the Kochen-Specker theorem.

    As I said, the crucial thing is to explain what do they mean by certain verbal expressions. By "physical" they mean "objectively real", and by "objectively real" they mean "basis independent". The last term "basis independent" is mathematically well defined, which allows them to prove rigorous theorems. The identification
    physical = objectively real = basis independent
    is merely a definition of the concepts, so they are neither right nor wrong about that. In their paper one simply needs to remember that otherwise vague terms "physical" and "objectively real" mean "basis independent", even if in some other papers these vague terms have a different meaning.

    Now let me express the content of their theorems in a more common language, in which the words physical and basis independent do not have the same meaning. With such a more common terminology, their theorems say that the physical content of QM is not basis independent. But this claim is not new at all. This is nothing but a restatement of the preferred basis problem appearing in one way or another in all interpretations of QM.

    The preferred basis problem is most explicit in the von Neumann collapse interpretation. If the wave function collapses due to observation, in what basis the collapse happens?

    Another example is the many-world interpretation. Even though there is no collapse, the preferred basis problem is a very serious one as discussed e.g. in https://www.physicsforums.com/threads/many-worlds-proved-inconsistent.767809/

    While it is already known that a preferred basis problem appears in all specific interpretations of QM, in each specific interpretation this problem takes a different form. By contrast, the theorem in the present paper does not assume any specific interpretation, thus presenting a universal interpretation-independent formulation of the preferred basis problem.

    This is similar to the status of contextuality in QM. The contextuality takes a different form in different interpretations of QM, while the Kochen-Specker theorem presents a universal interpretation-independent proof of contextuality.
     
    Last edited: Jan 5, 2015
  12. Jan 5, 2015 #11

    vanhees71

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    Quantum theory is in fact basis independent. It can be formulated without a basis, as was shown by Dirac in ~1926 and more mathematically rigorously by von Neumann. So I don't understand these statements at all. I also don't think that "collapse" should be part of any interpretation, at least not as a real physical process, because this makes more problems than anything else and is totally unnecessary.
     
  13. Jan 5, 2015 #12

    atyy

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    It is a matter of definition. If one chooses position to be real, then the real vectors are the eigenvectors of position. This picks out a preferred basis that is distinct from the momentum basis.
     
  14. Jan 5, 2015 #13

    vanhees71

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    But this is not a "preferred basis" but just the choice, which observable I want to measure. If I want to measure position, I'll take the generalized position eigenbasis to evaluate the position-probability distribution; if I want to measure momentum, I take the generalized momentum eigenbasis to evaluate the momentum-probability distribution.

    If I then filter according to a position or momentum range, I've prepared a new state. That's a physically meaningful preference but not a "preferred basis" in the sense as if there's a preferred basis in the theory as a whole.
     
  15. Jan 5, 2015 #14

    Demystifier

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    As long as you do not consider measurements, it is true that quantum theory is basis independent. But at that level, it is also physically empty. To give the physical meaning to the quantum theory, you must say something about what happens (or how the formalism has to be used) when a measurement is performed. So can you say something about that? And more importantly, can you say that in a basis independent way? Try to do it and I will tell you how this depends on the basis.
     
    Last edited: Jan 5, 2015
  16. Jan 5, 2015 #15

    atyy

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    Yes, you choose the preferred basis by your choice of measurement. It is just a matter of definition, not much different from saying that position and momentum cannot be simultaneously real.
     
  17. Jan 5, 2015 #16

    Demystifier

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    With such an operational view of quantum theory, the point is that it is you who is making the choice which observable to measure. The quantum state itself cannot make such a choice. So you are not the quantum state. But you are certainly physical, so there is something physical which is not a quantum state. Furthermore, as you are not a quantum state, you do not live in the Hilbert space and in that sense you cannot say that you are "basis independent". So there is something physical which is not basis independent.

    You might say that you are not basis dependent either, simply because you do not live in the Hilbert space implying that basis dependence/independence is simply not a property of you. That's correct, but there is still something basis dependent about you, because you choose one basis or another by choosing one observable or another to measure. A choice of an obserbable corresponds to a choice of a basis, because each observable defines a preferred basis - the one in which this observable is diagonal.
     
    Last edited: Jan 5, 2015
  18. Jan 5, 2015 #17

    atyy

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    Although their narrow conclusion is (probably) strictly correct, I wonder whether their wider discussion, such as their criticism of PBR etc is good or not. PBR uses a hidden variable framework, which includes Bohmian Mechanics. Their definition of real doesn't seem to impact at all whether the wave function is "real" in Bohmian Mechanics (where it can be considered real), and the basis dependence is (I think) just contextuality.
     
  19. Jan 5, 2015 #18

    Demystifier

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    Well, the point is that they use a different definition of "reality" than PBR (and Bohmians) do. As I said, they define "real" as "basis independent", which is not the definition of reality according to PBR or Bohmians. Thus, they are right that the state is "non-real" according to their definition, but PBR are also right that the state is "real" according to the PBR definition.
     
  20. Jan 5, 2015 #19

    vanhees71

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    According to quantum theory I am a quantum state, which behaves pretty classically, because I'm in exchange with the environment and consist of a large number of micro- states making up my macrostate, but that's another topic.

    The state of the quantum system is of course also choosen by me as an observer. If I'd be an experimentalist I'd even actively use some equipment to prepare quanta in a specific state, like the particle physicists at the LHC do when they prepare to proton beams banging head on at nearly 14 GeV center-mass energy pretty soon (hopefully). I don't see, where there should be a problem with a preferred basis. It's simply the state determination as described carefully, e.g., in Ballentine's book, i.e., within standard quantum theory in the minimal interpretation (which I also dubbed the "no-nonsense interperation" once :-)).
     
  21. Jan 5, 2015 #20

    atyy

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    Yes, I understood that their claim is probably right after reading your analysis in post #10. The interesting thing is they also say things like (p17) "As we have proven through the NDI theorem, the PBR theorem is derived from a false hypothesis making its result untenable." Presumably that is not a correct criticism of PBR then?
     
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